Wave Equation--Rectangle
المؤلف:
المرجع الالكتروني للمعلوماتيه
المصدر:
المرجع الالكتروني للمعلوماتيه
الجزء والصفحة:
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25-7-2018
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Wave Equation--Rectangle
To find the motion of a rectangular membrane with sides of length 
and 
(in the absence of gravity), use the two-dimensional wave equation
 |
(1)
|
where 
is the vertical displacement of a point on the membrane at position (
) and time 
. Use separation of variables to look for solutions of the form
 |
(2)
|
Plugging (2) into (1) gives
 |
(3)
|
where the partial derivatives have now become complete derivatives. Multiplying (3) by 
gives
 |
(4)
|
The left and right sides must both be equal to a constant, so we can separate the equation by writing the right side as
 |
(5)
|
This has solution
 |
(6)
|
Plugging (5) back into (◇),
 |
(7)
|
which we can rewrite as
 |
(8)
|
since the left and right sides again must both be equal to a constant. We can now separate out the 
equation
 |
(9)
|
where we have defined a new constant 
satisfying
 |
(10)
|
Equations (◇) and (◇) have solutions
 |
(11)
|
 |
(12)
|
We now apply the boundary conditions to (11) and (12). The conditions 
and 
mean that
 |
(13)
|
Similarly, the conditions 
and 
give 
and 
, so 
and 
, where 
and 
are integers. Solving for the allowed values of 
and 
then gives
 |
(14)
|
Plugging (◇), (◇), (◇), (◇), and (14) back into (◇) gives the solution for particular values of 
and 
,
 |
(15)
|
Lumping the constants together by writing 
(we can do this since 
is a function of 
and 
, so 
can be written as 
) and 
, we obtain
 |
(16)
|
Plots of the spatial part for modes are illustrated above.
The general solution is a sum over all possible values of 
and 
, so the final solution is
 |
(17)
|
where 
is defined by combining (◇) and (◇) to yield
 |
(18)
|
Given the initial conditions 
and 
, we can compute the 
s and 
s explicitly. To accomplish this, we make use of the orthogonality of the sine function in the form
 |
(19)
|
where 
is the Kronecker delta. This can be demonstrated by direct integration. Let 
so 
in (◇), then
 |
(20)
|
Now use the trigonometric identity
 |
(21)
|
to write
![I=L/(2pi)int_0^picos[(m-n)u]du+int_0^picos[(m+n)u]du.](http://mathworld.wolfram.com/images/equations/WaveEquationRectangle/NumberedEquation22.gif) |
(22)
|
Note that for an integer 
, the following integral vanishes
since 
when 
is an integer. Therefore, 
when 
. However, 
does not vanish when 
, since
 |
(27)
|
We therefore have that 
, so we have derived (◇). Now we multiply 
by two sine terms and integrate between 0 and 
and between 0 and 
,
 |
(28)
|
Now plug in 
, set 
, and prime the indices to distinguish them from the 
and 
in (28),
 |
(29)
|
Making use of (◇) in (29),
 |
(30)
|
so the sums over 
and 
collapse to a single term
 |
(31)
|
Equating (30) and (31) and solving for 
then gives
 |
(32)
|
An analogous derivation gives the 
s as
 |
(33)
|
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